# unit 1 assessment project skylar larson. 0.00 seconds 1.50 seconds1.00 seconds 0.50 seconds 3.50...

Post on 03-Jan-2016

219 views

Category:

## Documents

Tags:

• #### negative infinity

Embed Size (px)

TRANSCRIPT

Particle Motion Project

Unit 1 assessment projectSkylar Larson

0.00 seconds1.50 seconds1.00 seconds0.50 seconds3.50 seconds3.00 seconds2.50 seconds2.00 seconds

6.01905x^2-5.4x+3.55950.0 < x < 0.50-12.496696x^2+43.852139x-21.19176580.50 < x < 2.875-4.3429x^2+36.62286x-68.632.875 < x < 3.50f(x)=Edisplacement equations

Height vs. Time GraphVelocity equationsf(x)=-24.99339x+43.85213912.0381x-5.4 0.50 < x < 2.875-8.6858x+36.62286 2.875 < x < 3.500.0 < x < 0.50Velocity vs. Time Graph

Average Velocity T=0sT=3.5s

3.5ft

6.2ftAverage Velocity= (6.2 - 3.5)/(3.5 - 0)

= 0.77143 ft/sInstantaneous Velocity at t=2.0secondsf(2) = -6.135 ft/sf(x)= -24.99339x+43.852139 0.50 < x < 2.875 2.0 seconds (From preview page)Instantaneous velocity at T=3.5 secondsf(x)= 6.2226 ft/s f(x)= -8.6858x+36.622862.875 < x < 3.50(From preview page)3.5 seconds Did the ball ever travel at 5 m/s (16.404 ft/s)?12.0381x-5.4 = 16.4040.0 < x < 0.50x=1.81125

-24.99339x+43.852139 = 16.404 0.50 < x < 2.875 x=1.09822-8.6858x+36.62286 =16.404 2.875 < x < 3.50 x=2.3278f(x)= The ball will reach 5m/s (16.404 ft/s) at 1.09822 seconds. The other two x values are not in the domain. Part 2

Definition of Derivative: lim = (f(x+h) f(x-h))/(2h) h 0

ft/sft/sInstantaneous rate of change at 2.0625 seconds2.0625 seconds

Part 3f(x)=-2x+4 -1 < x < 1-(x-1)^2+2 1 < x < 4-0.5|x-8|+6 4 < x < 12Not continuous at x=4 because lim f(x)=4

And lim f(x)=7

x 4+x 4-They are not equalPart 3

f(x)=-2x+4 -1 < x < 1-(x-1)^2+2 1 < x < 4-0.5|x-8|-5 4 < x < 12Change function so there is a limit: move the absolute value equation/line down 11 units.Lim f(x) = -7 x 4Lim (x)= -7 x 4+-The limit does not exist at x = 4 because the left and right limits dont equal each other.Part 4Has a limit approaching infinity, as x approaches infinity.F(x)= (3x^2+4x)/(2x+7)Horizontal Asymptotes: None

Part 4Has a limit approaching 0, as x approaches infinity.

F(x)= (7x^2+3)/(2x^4+x)Horizontal Asymptotes: y=0 Part 4Has a limit approaching a line which is not 0, as x approaches infinity.

F(x)= (3x^2+4x)/(4x^2)Horizontal Asymptotes: y= 0.75 (Found using coefficients)Part 4Has a limit approaching two separate lines as x approaches positive or negative infinity.

F(x)= (|2x|)/(3x)Horizontal Asymptotes: y=2/3 and y= -2/3 (Found using coefficients)

Recommended